WASHINGTON FUNDAMENTALS OF ENGINEERING
(FE) PRACTICE EXAMINATION STUDY GUIDE |
LATEST UPDATE 2026/2027 | ACTUAL EXAM |
PRACTICE QUESTIONS AND ANSWERS | EXAM
REVIEW | 100% CORRECT ANSWERS | VERIFIED
SOLUTIONS
This practice examination is designed for candidates preparing for the NCEES
Fundamentals of Engineering (FE) examination for professional engineering
licensure in Washington State. The FE examination is a comprehensive,
computer-based test that assesses knowledge of fundamental engineering
principles and practices across all disciplines. This 100-question practice exam
covers the core topics outlined in the NCEES FE Reference Handbook, including
mathematics, probability and statistics, engineering economics, ethics and
professional practice, statics, dynamics, mechanics of materials, fluid mechanics,
thermodynamics, materials science, and electricity and magnetism. Each question
is crafted to mirror the difficulty and format of the actual exam, integrating
conceptual understanding with analytical problem-solving. Detailed,
multi-sentence rationales accompany each answer to reinforce learning and
identify knowledge gaps. Updated for the 2026–2027 examination cycle, this
resource provides verified solutions to help you build confidence and successfully
pass the FE exam on your first attempt.
Table of Contents
I. Mathematics and Analytical Geometry
II. Probability and Statistics
III. Engineering Economics
IV. Ethics and Professional Practice
V. Statics
VI. Dynamics
VII. Mechanics of Materials
VIII. Fluid Mechanics
, IX. Thermodynamics and Heat Transfer
X. Materials Science and Engineering
XI. Electricity and Magnetism
1. A particle moves along a straight line with a velocity given by v(t) = 2t – 4,
where v is in meters per second and t is in seconds. What is the total
distance traveled by the particle between t = 0 and t = 4 seconds?
A) 0 m
B) 6 m
C) 8 m
D) 10 m
Correct Answer: C
The velocity function v(t) = 2t – 4 is linear and changes sign at t = 2 s, indicating a
reversal of direction. From t = 0 to t = 2 s, velocity is negative, and the particle
moves backward; the displacement is ∫₀² (2t – 4) dt = [t² – 4t]₀² = (4 – 8) = –4 m,
giving a distance of 4 m. From t = 2 to t = 4 s, velocity is positive, and the
displacement is ∫₂⁴ (2t – 4) dt = [t² – 4t]₂⁴ = (16 – 16) – (4 – 8) = 0 – (–4) = 4 m,
distance 4 m. The total distance traveled is the sum of absolute distances, 4 m + 4
m = 8 m. Options A and B are incorrect because they fail to account for the
direction change or miscalculate the integral; D overestimates the distance. Thus,
C is correct.
2. Evaluate the definite integral ∫₀² (3x² + 2x) dx.
A) 10
B) 12
C) 14
D) 16
Correct Answer: B
The antiderivative of 3x² + 2x is x³ + x². Applying the limits, F(2) = (2)³ + (2)² = 8 + 4
,= 12, and F(0) = 0. The definite integral equals 12 – 0 = 12. Option A is too low, C
and D are incorrect calculations. Thus, B is correct.
3. The determinant of the matrix A = [[2, 3], [4, 5]] is:
A) –2
B) 2
C) 7
D) 22
Correct Answer: A
For a 2×2 matrix [[a, b], [c, d]], the determinant is ad – bc. Here, a = 2, b = 3, c = 4,
d = 5, so the determinant is (2)(5) – (3)(4) = 10 – 12 = –2. Option B is the absolute
value, C is a + d, and D is the product of diagonals summed incorrectly. Therefore,
A is correct.
4. In a sample of 100 bolts, the mean tensile strength is 50 ksi with a standard
deviation of 5 ksi. What is the 95% confidence interval for the population
mean, assuming a z-value of 1.96?
A) 49.0 to 51.0 ksi
B) 49.5 to 50.5 ksi
C) 48.0 to 52.0 ksi
D) 40.0 to 60.0 ksi
Correct Answer: A
The confidence interval is given by mean ± z × (standard deviation / √n). Here, 50 ±
1.96 × (5 / √100) = 50 ± 1.96 × 0.5 = 50 ± 0.98, resulting in approximately 49.02 to
50.98 ksi, which rounds to 49.0 to 51.0 ksi. Option B uses an incorrect divisor, C
uses a z of 2 with wider range, and D is far too wide. Thus, A is correct.
5. What is the probability of drawing two aces in succession from a standard
deck of 52 cards without replacement?
A) 1/221
B) 1/169
C) 4/663
D) 1/13
, Correct Answer: A
The probability of the first ace is 4/52 = 1/13. After drawing one ace, 3 aces
remain out of 51 cards, so the probability of the second ace is 3/51 = 1/17. The
combined probability is (1/13) × (1/17) = 1/221. Option B assumes replacement, C
is 4/52 × 4/51, and D is the probability of a single ace. Therefore, A is correct.
6. A machine has an initial cost of $10,000, an annual operating cost of
$2,000, and a salvage value of $3,000 after 5 years. Using an interest rate of
5%, what is the present worth of the machine's total cost?
A) –$16,220
B) –$14,580
C) –$18,760
D) –$12,940
Correct Answer: A
The present worth is the initial cost plus the present worth of the annual operating
costs minus the present worth of the salvage value. Annual operating cost present
worth = $2,000 × (P/A, 5%, 5) = $2,000 × 4.3295 = $8,659. Salvage present worth =
$3,000 × (P/F, 5%, 5) = $3,000 × 0.7835 = $2,350.50. Total present worth = –
$10,000 – $8,659 + $2,350.50 = –$16,308.50, approximately –$16,220 after
rounding differences in factors. Option B omits salvage, C adds salvage instead of
subtracting, D uses incorrect factors. Thus, A is the closest correct value.
7. According to the NCEES Model Rules of Professional Conduct, an engineer
who discovers a client's violation of environmental regulations should first:
A) Report the violation immediately to the appropriate regulatory agency
B) Advise the client of the violation and recommend corrective action
C) Withhold further services until the client pays additional fees
D) Ignore the violation if it does not directly affect public safety
Correct Answer: B
The NCEES Model Rules require an engineer to first advise the client or employer of
any observed violation and recommend corrective measures. If the client fails to
act, the engineer may then report to appropriate authorities. Option A skips the
(FE) PRACTICE EXAMINATION STUDY GUIDE |
LATEST UPDATE 2026/2027 | ACTUAL EXAM |
PRACTICE QUESTIONS AND ANSWERS | EXAM
REVIEW | 100% CORRECT ANSWERS | VERIFIED
SOLUTIONS
This practice examination is designed for candidates preparing for the NCEES
Fundamentals of Engineering (FE) examination for professional engineering
licensure in Washington State. The FE examination is a comprehensive,
computer-based test that assesses knowledge of fundamental engineering
principles and practices across all disciplines. This 100-question practice exam
covers the core topics outlined in the NCEES FE Reference Handbook, including
mathematics, probability and statistics, engineering economics, ethics and
professional practice, statics, dynamics, mechanics of materials, fluid mechanics,
thermodynamics, materials science, and electricity and magnetism. Each question
is crafted to mirror the difficulty and format of the actual exam, integrating
conceptual understanding with analytical problem-solving. Detailed,
multi-sentence rationales accompany each answer to reinforce learning and
identify knowledge gaps. Updated for the 2026–2027 examination cycle, this
resource provides verified solutions to help you build confidence and successfully
pass the FE exam on your first attempt.
Table of Contents
I. Mathematics and Analytical Geometry
II. Probability and Statistics
III. Engineering Economics
IV. Ethics and Professional Practice
V. Statics
VI. Dynamics
VII. Mechanics of Materials
VIII. Fluid Mechanics
, IX. Thermodynamics and Heat Transfer
X. Materials Science and Engineering
XI. Electricity and Magnetism
1. A particle moves along a straight line with a velocity given by v(t) = 2t – 4,
where v is in meters per second and t is in seconds. What is the total
distance traveled by the particle between t = 0 and t = 4 seconds?
A) 0 m
B) 6 m
C) 8 m
D) 10 m
Correct Answer: C
The velocity function v(t) = 2t – 4 is linear and changes sign at t = 2 s, indicating a
reversal of direction. From t = 0 to t = 2 s, velocity is negative, and the particle
moves backward; the displacement is ∫₀² (2t – 4) dt = [t² – 4t]₀² = (4 – 8) = –4 m,
giving a distance of 4 m. From t = 2 to t = 4 s, velocity is positive, and the
displacement is ∫₂⁴ (2t – 4) dt = [t² – 4t]₂⁴ = (16 – 16) – (4 – 8) = 0 – (–4) = 4 m,
distance 4 m. The total distance traveled is the sum of absolute distances, 4 m + 4
m = 8 m. Options A and B are incorrect because they fail to account for the
direction change or miscalculate the integral; D overestimates the distance. Thus,
C is correct.
2. Evaluate the definite integral ∫₀² (3x² + 2x) dx.
A) 10
B) 12
C) 14
D) 16
Correct Answer: B
The antiderivative of 3x² + 2x is x³ + x². Applying the limits, F(2) = (2)³ + (2)² = 8 + 4
,= 12, and F(0) = 0. The definite integral equals 12 – 0 = 12. Option A is too low, C
and D are incorrect calculations. Thus, B is correct.
3. The determinant of the matrix A = [[2, 3], [4, 5]] is:
A) –2
B) 2
C) 7
D) 22
Correct Answer: A
For a 2×2 matrix [[a, b], [c, d]], the determinant is ad – bc. Here, a = 2, b = 3, c = 4,
d = 5, so the determinant is (2)(5) – (3)(4) = 10 – 12 = –2. Option B is the absolute
value, C is a + d, and D is the product of diagonals summed incorrectly. Therefore,
A is correct.
4. In a sample of 100 bolts, the mean tensile strength is 50 ksi with a standard
deviation of 5 ksi. What is the 95% confidence interval for the population
mean, assuming a z-value of 1.96?
A) 49.0 to 51.0 ksi
B) 49.5 to 50.5 ksi
C) 48.0 to 52.0 ksi
D) 40.0 to 60.0 ksi
Correct Answer: A
The confidence interval is given by mean ± z × (standard deviation / √n). Here, 50 ±
1.96 × (5 / √100) = 50 ± 1.96 × 0.5 = 50 ± 0.98, resulting in approximately 49.02 to
50.98 ksi, which rounds to 49.0 to 51.0 ksi. Option B uses an incorrect divisor, C
uses a z of 2 with wider range, and D is far too wide. Thus, A is correct.
5. What is the probability of drawing two aces in succession from a standard
deck of 52 cards without replacement?
A) 1/221
B) 1/169
C) 4/663
D) 1/13
, Correct Answer: A
The probability of the first ace is 4/52 = 1/13. After drawing one ace, 3 aces
remain out of 51 cards, so the probability of the second ace is 3/51 = 1/17. The
combined probability is (1/13) × (1/17) = 1/221. Option B assumes replacement, C
is 4/52 × 4/51, and D is the probability of a single ace. Therefore, A is correct.
6. A machine has an initial cost of $10,000, an annual operating cost of
$2,000, and a salvage value of $3,000 after 5 years. Using an interest rate of
5%, what is the present worth of the machine's total cost?
A) –$16,220
B) –$14,580
C) –$18,760
D) –$12,940
Correct Answer: A
The present worth is the initial cost plus the present worth of the annual operating
costs minus the present worth of the salvage value. Annual operating cost present
worth = $2,000 × (P/A, 5%, 5) = $2,000 × 4.3295 = $8,659. Salvage present worth =
$3,000 × (P/F, 5%, 5) = $3,000 × 0.7835 = $2,350.50. Total present worth = –
$10,000 – $8,659 + $2,350.50 = –$16,308.50, approximately –$16,220 after
rounding differences in factors. Option B omits salvage, C adds salvage instead of
subtracting, D uses incorrect factors. Thus, A is the closest correct value.
7. According to the NCEES Model Rules of Professional Conduct, an engineer
who discovers a client's violation of environmental regulations should first:
A) Report the violation immediately to the appropriate regulatory agency
B) Advise the client of the violation and recommend corrective action
C) Withhold further services until the client pays additional fees
D) Ignore the violation if it does not directly affect public safety
Correct Answer: B
The NCEES Model Rules require an engineer to first advise the client or employer of
any observed violation and recommend corrective measures. If the client fails to
act, the engineer may then report to appropriate authorities. Option A skips the
